A car traveling at 50 km/h hits a bridge abutment. A passenger in the car moves forward a distance of 61 cm (with respect to the road) while being brought to rest by an inflated air bag. What magnitude of force (assumed constant) acts on the passenger's upper torso, which has a mass of 44 kg?

6957.04N

Explanation:

Using

vf2=vi2+2ad

But vf = 0 .

So convert 50km/hr to m/s, and you need to convert 61 cmto m

(50km/hr)*(1hr/3600s)*(1000m/km) = 13.9m/s

61cm * (1m/100cm) = .61m

So n

0 = (13.9m/s)^2 + 2a(.61m)

a = 158.11m/s^2

So

using F = ma

F = 44kg(158.11m/s^2) = 6957.04N

6.95 kN

Explanation:

Given that

Speed of the car, u = 50 kmph

Distance moved by the passenger, s = 61 cm = 0.061 m

Mass of the passenger, m = 44 kg

Converting the initial velocity from km/h to m/s, we have

50 kmph = 50 * 1000/3600

50 kmph = 13.89 m/s

Using one of the equations of motion,

v² = u² + 2as, where v = 0,making a the subject of formula

a = -u²/2s

a = -(13.89²) / 2 * 0.61

a = -192.93 / 1.22

a = -158 m/s²

The force acting on the passenger's leg, F = m.a, so

-F = 44 * -158

F = 6952 N, or 6.95 kN

c)as speed changes, wavelength and frequency change.

everything changes!

i think it would be 5